So |a + b| < |a| + |b|. Proved. |a+b|<=|a|+|b|?

[Agent M27]

Homework Statement

Let a & b be real numbers.

Prove that:
|a+b|<=|a|+|b|

Homework Equations

|x|=$$\sqrt{x^{2}}$$

The Attempt at a Solution

|a+b|

=$$\sqrt{(a+b)^{2}}$$

=$$\sqrt{(a^{2}+2ab+b^{2})}$$ <= [tex]\sqrt{a^{2}} + $$\sqrt{b^{2}}$$<br /> <br /> |a|=$$\sqrt{a^{2}}$$<br /> <br /> |b|=$$\sqrt{b^{2}}$$<br /> <br /> I feel like this is lacking in foundation, but I lack in the foundation of proofs involving absolute value. Thanks in advance for the assistance.<br /> <br /> Joe<br /> <br /> Sorry for the ugly formatting, tex is cumbersome sometimes.[/tex]

 

[micromass]

This is ok. But I would like some more details in the following inequality:

$$\sqrt{a^2+2ab+b^2}\leq\sqrt{a^2}+\sqrt{b^2}$$

It's not immediately obvious why this should be true...

 

[HallsofIvy]

Homework Statement

Let a & b be real numbers.

Prove that:
|a+b|<=|a|+|b|

Homework Equations

|x|=$$\sqrt{x^{2}}$$

The Attempt at a Solution

|a+b|

=$$\sqrt{(a+b)^{2}}$$

=$$\sqrt{(a^{2}+2ab+b^{2})}$$ <= $$\sqrt{a^{2}} + \sqrt{b^{2}}$$

This is the only place I see a difficulty. How do you prove this? I suspect it is no simpler to prove this than it is to prove $|a+ b|\le |a|+ |b|$ by "considering cases": a positive or negative, b positive or negative.

|a|=$$\sqrt{a^{2}}$$

|b|=$$\sqrt{b^{2}}$$

I feel like this is lacking in foundation, but I lack in the foundation of proofs involving absolute value. Thanks in advance for the assistance.

Joe

Sorry for the ugly formatting, tex is cumbersome sometimes.

 

[╔(σ_σ)╝]

This is ok. But I would like some more details in the following inequality:

$$\sqrt{a^2+2ab+b^2}\leq\sqrt{a^2}+\sqrt{b^2}$$

It's not immediately obvious why this should be true...

I completely agree. In fact, I have never seen this before.

How can you prove this ?

 

[chiro]

Homework Statement

Let a & b be real numbers.

Prove that:
|a+b|<=|a|+|b|

Homework Equations

|x|=$$\sqrt{x^{2}}$$

The Attempt at a Solution

|a+b|

=$$\sqrt{(a+b)^{2}}$$

=$$\sqrt{(a^{2}+2ab+b^{2})}$$ <= [tex]\sqrt{a^{2}} + $$\sqrt{b^{2}}$$<br /> <br /> |a|=$$\sqrt{a^{2}}$$<br /> <br /> |b|=$$\sqrt{b^{2}}$$<br /> <br /> I feel like this is lacking in foundation, but I lack in the foundation of proofs involving absolute value. Thanks in advance for the assistance.<br /> <br /> Joe<br /> <br /> Sorry for the ugly formatting, tex is cumbersome sometimes.[/tex]

$$<br /> With vectors the best way is to use property of the inner product.<br /> <br /> For this however I would simply consider the cases of signs. Because these are numbers in one dimension, you only have to check a few cases.<br /> <br /> If sgn(a) = sign(b) then |a + b| = |a| + |b| if a,b >= 0. |a + b| = |-c - d| = |-(c+d)|<br /> = |-c| + |-d|<br /> <br /> Now let sgn(a) = 1 - sgn(b)<br /> <br /> Then |a + b| < |a| + |b| can be proved simply from the property<br /> <br /> |a + b|^2 = a^2 + b^2 + 2ab<br /> <br /> But if sgn(a) = 1 - sgn(b) then 2ab < 0<br /> <br /> So |a+b|^2 < |a|^2 + |b|^2 from above property$$